new approach to solving generalised linear goal ... · new approach to solving generalised linear...

Mathematics Applied in Science and Technology.

ISSN 0973-6344 Volume 9, Number 1 (2017) pp 1-22

© Research India Publications

http://www.ripublication.com

New Approach to Solving Generalised Linear Goal

Programming Problem

Orumie, Ukamaka Cynthia

Lecturer Department of Mathematics and Statistics,

University Of Port Harcourt, Nigeria

Abstract

A new approach to solving any \linear goal programming problems is developed

irrespective of whether it is in lexicographic, weighted or generalized form. Also

the method by Orumie and Ebong 2013a and 2013b were modified to incorporate

the rigid constraints in the initial table. Illustrative example were given, and the

new algorithm proved better.

Keyword: Algorithm, goal programming, Lexicographic, Weighted,

INTRODUCTION

Since the development of Goal Programming, it’s study has been an important topic

of interest for researchers in the area of multi-criteria decision making. In a view to

solving especially Linear GPPs, the most commonly used goal programming solution

methods were introduced by Lee (1972) and Ignizio (1976) ( see Schniederjans and

Kwak(1982)). According to Schniederjans and Kwak (1982), these algorithms

consume a lot of time and space during computation involving a large number of

constraints (Orumie and Ebong (2014)).

2 Orumie, Ukamaka Cynthia

In 2011, Orumie and Ebong developed an algorithm and compared it with the most

commonly used ones. In (2013a) and (2013b), Orumie and Ebong developed separate

methods of solving lexicographic and weighted goal programming respectively which

corrected most of the computational errors in other existing algorithm as highlighted

in Orumie and Ebong (2014). This new algorithm corrected the error in the alternative

method of solving Linear goal programming problem model by Orumie and Ebong

(2011), with inclusion of the rigid constraints.

However, the new algorithms in testing for optimality did not consider the fact that

there exist a situation where absolute value of coefficient of deviational variable under

consideration that are not in the table can re-enter. Also in case of choosing the

enterying variable , the algorithm did not consider in case of ties in the minimum of

the ratio of the right hand side to the positive coefficients of the columns

corresponding to the ties. The initial table of both Orumie and Ebong 2011 and 2013

did not include rigid constraints.

Thus, a major drawback in the use of goal programming problems has been the lack

of an algorithm, capable of reaching optimal solution in reasonable time (see Olson

(1984)).

In order to reduce the computational time of GPP, and correct the above errors,

the researcher decided to develop an algorithm for linear goal programming that can

minimise computational time when compared with existing algorithms irrespective of

its structure. The algorithm could be used to solve problems represented either in pre-

emptive model, weighted model, or prioritized (generalized or combined models).

Section two shows several kinds of linear goal programming structure. The initial

table is formulated and presented in section three. Section four, five, and six shows

the modified lexicographic, modified weighted and the new generalized method

respectively. Section seven is the illustrative example wheras section eight and nine is

summary and conclusion.

KINDS OF LINEAR GOAL PROGRAMMING REPRESENTATION

Consider the Linear Goal programming model formulation for n variables, r goals

(constraints), s rigid constraints, and K preemptive priority factors defined below:

Find x = ( nxxx ...,,, 21 ) that minimizes

z = { L1(d - , d+), L2(d - , d+), L3(d - , d+), …, LK(d - , d+)} (1)

subject to

New Approach to Solving Generalised Linear Goal Programming Problem 3

fi( x )+di- -di

+ = bi for all i = 1,... ,r (2)

gi( x ) hi for all i=1, 2, . . ., (3)

and x 0, d- 0, d+ 0 (4)

Here, xj is the jth decision variable;

z represents the objective function which measures the achievement of the goals at

each priority level k;

d - = (d1- ,d2

- , ...,dr- ) , d+ =(d1

+ , d2+ ,...,dr

+ ) ;

di- is the ith negative deviational variable;

id is the ith positive deviational variable;

Lk is kth linear function of the deviational variables associated with objectives at

priority level k, k= 1,2, ..., K;

K is the total number of priority factors in the model,

bi is the right-hand side constant for goal constraint i;

hi is the right-hand side constant for rigid constraint i;

fi

is the linear function on the left hand side of the ith goal constraint;

gi

is the linear function on the left hand side of the ith rigid constraint;

let m =r+

Then, examples of linear goal programming structures will be enumerated to highlight

the differences between the proposed and existing algorithms.

The first example (example 1) given below is one where the objective function is

strictly lexicographic.

Example .1

Min z = p1 d2- , p2 d1

- , p3

3d

120 x1 + 90 x2 + d1- - d1

+ = 1950

x1 + d2-- d2

+ = 13

x2 + d3 - - d3

+ = 5

6 x1+ 3 x2 90

3 x1 + 6 x2 72


Table 1: Initial table of problem .1 of new algorithm

Var in z with

priority

x1 x2

1d d2

- d3- s1 s2 RHS

p2 d1- 120 90 1 0 0 0 0 1950

p1d2- 1 0 0 1 0 0 0 13

p3d3- 0 1 0 0 1 0 0 5

s1 6 3 0 0 0 1 0 90

s2 3 6 0 0 0 0 1 72

Column one consists of the deviational variables that appeared in the objective with

priority assigned to each of them, together with slack variables from the rigid

constraints. These form the basis. All other columns excepts the last column forms

coefficients of the decision variables from both goal and rigid constraints, deviational

variables that appeared in z, and slack variables from the rigid constraints as appeared

in constraint equation.

The first column of row two is the deviational variable of the first goal constraint that

appears in the objective function. This corresponds to priority level two (p2) ,and the

rest of the entries of row two are the coefficients of decision variables from both goal

and rigid constraints, deviational variables that appeared in z, and slack variables from

the rigid constraints as appeared in constraint equation. Similarly, on the 3rd row of

column one refers to the deviational variable of the 2nd goal constraint that appeared

in the objective function. This corresponds to priority p1, and the rest of the entries of

row three follows from that of row two. Row four of column one refers to the

deviational variable from 3rd goal constraint that appeared in the objective function,

which corresponds to the 3rd priority p3. The last two entries on the first column of the

last two rows refer to the respective slacks of the rigid constraints.


REMARKS

The new approach utilizes only the deviational variables that appeared in z on both

basis and non basis, and does not include objective function rows. Instead, priority is

attached to the deviational variables in column one and any other deviational variable

column augmented if need be. This tpe of problem can be handled by Orumie and

Ebong (2013) but its initial table omitted the slack and surplus rows from rigid

constraints.

Another example (example 2) to be represented in a table is a case where the

objective function is a weighted sum of the deviational variables (Weighted GP).

Example 2

12(min dz +4d2

+)

s.t

121 84 dxx -

1d =45

8x1 + 24x2 +

2d –

2d = 100

x1 + 2x2 ≤ 10

x1 ≤ 9

Therefore, inital table of example .2 using new algorithm is represented below;

Table 2: Inital Table of example 2 using new algorithm

Variable in basis as

appeared in z with si .

x1 x2 d1- d2

+ s1 s2 Solution value

bi (rhs)

2 d1-

4d2+

4 8 1 0 0 0 45

8 24 0 0 0 0 100

s1

s2

1 2 0 0 1 0 10

1 0 0 0 0 1 9

Column one consists of the deviational variables that appeared in z with weight

assigned to each of them, and slack variables from the rigid constraints which forms

the basis. The rest of the entries are obtained in manner similar to that used to obtain

the entries in example 1


The third example (example 3) given below is one where both weighted and

preemptive approaches are combined to form a model to a problem. (i.e goals

categorised into groups, where the goals within each group are of equal importance,

but there are slight differences between the groups in their level of importance such

that weight is used within each group)(Generalised variant).

Example 3

min z = p1 (d1-+ 3d2

- +2 d3- ), p2(d4

+), p3(d5+)

s.t

121 dxx -

1d =30

x3 +x4 +

2d –

2d =30

3x1 + 2x3 +d3- - d3

+ =120

3x2 + 2x4 +d4- - d4

+ =20

10 x1 +9x2 + 8 x3 +7x4 + d5- -d5

+=800

6 x1+ 3 x2 90,

3 x1 + 6 x2 72

x1, x2, x3, x4, d1-, d1

+, d2-, d2

+, d3-, d3

+ , d4- ,d4

+ , d5- ,d5

+ > 0

Table 3: Initial table of the proposed method on example 3

Variable in basis

as appeared

in z with si .

x1 x2 x3 x4 d1- d2

- d3- d4

+ d5+ s1 s2 Solution

value bi (rhs)

p1 d1- 1 1 0 0 1 0 0 0 0 0 0 30

3p1 d2- 0 0 1 1 0 1 0 0 0 0 0 30

2p1 d3- 3 0 2 0 0 0 1 0 0 0 0 120

p2d4+ 0 3 0 2 0 0 0 -1 0 0 0 20

p3 d5+

10 9 8 7 0 0 0 0 -1 0 0

800

s1 6 3 0 0 0 0 0 0 0 1 0 90

s2 3 6 0 0 0 0 0 0 0 0 1 72


Column one is the basis which consists of the deviational variables in the objective

function with priority and weight assigned to each of them, and slack variables from

the rigid constraints in order at which they appeared in constraint equations.

Column one of row two is the deviational variable of the first goal constraint that

appears in the objective function. Similarly, on the 3rd row of column one refers to the

deviational variable of the 2nd goal constraint that appeared in the objective function

with weight 3 attached to it. Row four of column one refers to the deviational variable

from 3rd goal constraint that appeared in the objective function with weight 2 attached

to it. These three rows correspond to priority p1. The rest of the entries are obtained in

manner similar to that used to obtain the entries in previous examples.

The above problem can be solved using Orumie and Ebong (2011), but its initial

table did not consider the slack and surplus rows from rigid constraints.

Another example (example 4) is a generalised case where both weighted and

preemptive approaches are combined to form a model to a problem, and also d- and d+

(positive and negative deviational variables) from the same constraint equation are in

the achievement function with different priorities assigned to each of them.

Example 3.4

min z = p1 (d1-+ 3d2

- +2 d3- ), p2(d4

+), p3(d5+), p4d1

+

s.t

x1+ x2 +

1d -

1d =30

x3 +x4 +

2d –

2d =30

3x1 + 2x3 +d3- - d3

+ =120

3x2 + 2x4 +d4- - d4

+ =20

10 x1 +9x2 + 8 x3 +7x4 + d5- -d5

+=800

6 x1+ 3 x2 90,

3 x1 + 6 x2 72

x1, x2, x3, x4, d1-, d1

+, d2-, d2

+, d3-, d3

+ , d4- ,d4

+ , d5- ,d5

+ > 0

The initial table of the proposed method of solving the same example is given below;


Table 4: Innitial table of example 4 for the proposed method

Variable in

basis as

appeared

in z with si .

x1 x2 x3 x4 d1- d2

- d3- d4

+ d5+ d1

+ s1 s2

Solution

value

bi (rhs)

p1d1- 1 1 0 0 1 0 0 0 0 -1 0 0 30

3p1 d2- 0 0 1 1 0 1 0 0 0 0 0 0 30

2p1 d3- 3 0 2 0 0 0 1 0 0 0 0 0 120

p2d4+ 0 3 0 2 0 0 0 -1 0 0 0 0 20

p3 d5+ 10 9 8 7 0 0 0 0 -1 0 0 0 800

s1 6 3 0 0 0 0 0 0 0 0 1 0 90

s2 3 6 0 0 0 0 0 0 0 0 0 0 72

In example 4 above, it is observed that both d1- and d1

+ appeared in the objective

function with different priorities attached to them. In the new method, if both d- and d+

(negative and positive deviational variables) from the same goal constraint are in the

achievement function but with different priorities, then the one with higher priority

will be in the basis whereas the lesser one will be placed alongside with other

variables in the non basis. This situation was not considered in the previous methods.

Therefore, column one of row two is the deviational variable (d1-) of the first goal

constraint that appears in the objective function which corresponds to priority p1. The

variable d1+ which corresponds to p4 appears only in the non-basis, since the one with

higher priority d1- has been represented in basis. The rest of the entries are obtained in

manner similar to that used to obtain the entries in previous examples.

INITIAL TABLE FOR THE GENERAL LINEAR GOAL PROGRAMMING

We now present the initial table of the general Linear GP using the guidelines we

develop for our proposed method. The general linear GP we will consider is the

following;


Find x =( nxxx .,..,, 21 ) that minimizes

z = { L1(d - , d+), L2(d - , d+), L3(d - , d+), …, LK(d - , d+)} (5)

subject to

n

jjij xa +di

- -di+ = bi for all i = 1,... ,r (6)

n

jjij xc hi for all i=1, 2, . . ., (7)

and xj 0, d - 0, d+ 0 (8)

where

n

jjij xa

is the linear function on the left hand side of the ith goal constraint;

n

jjij xc is the linear function on the left hand side of the ith rigid constraint;

The initial table of the above problem (formulation) is given in table 5 below. In the

table, the basic variables are determined in the manner illustrated in the above

examples. The ivid are the positive (

id ) and negative (

id ) deviational variables. If i

=1, then - i = -1 and if i =-1, then - i =1 in the table.

Table 5: Innitial table of the New Algorithms on LGP

Variable in

basis as

appeared

in z with si .

x1 x2 … xn 1

1

vd 2

2

vd 3

3

vd . . .rv

rd s1 s2 . . . s Solution

value (rhs)

Lkkv

kd

a11 a12 … a1n - 1 0 0 . . .0 0 0 . . . 0 b1

a21 a22 … a2n 0 - 2 0 . . . 0 0 0 . . . 0 b2

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

ar1 ar2 … arn 0 0 0 . .. - r 0 0 ... 0

.

. .

br

s1

s2

. .

c11 c12 … c1n 0 0 0 . . . 0 1 0 . .. 0 h1

c21 c22 … c2n 0 0 0 . . . 0 0 1 ... 0 h2

. . . . . . . . . . .


s .

. . . . . . . . . . .

. . . . . . . . . .

1c 2c … nc 0 0 0 . . . 0 0 0 . . . 1

.

.

h

Lk kvkd ; k= 1,2,…,r in row 2 column 1 in table 5 are obtained as illustrated in the

examples above.

The solution procedure for the above problem in table 5 considers the goal constraints

(gk.) as both objective functions and the constraints, where gk. is the kth priority row(s)

. That is, there are no inclusions of cj-zj rows on the table. It starts by not including

the deviational variables column that did not appear in the achievement function in

the tableau while searching for the optimal solution, but can be augmented when

necessary. This is because, positive deviational variables columns coefficient is the

same as negative of the negative deviational variable column coefficient, that is di-= -

di+. In this new method, the algorithm starts with the highest priority row constraints

and terminates with the lowest priority. Solution is obtained provided that for all pk

the steps in the algorithm are strictly maintained as shown in section 4 below.

MATERIALS AND METHOD (THE MODIFIED ALGORITHM FOR THE

LGP MODEL)

Let pk be the thk priority assigned to deviational variable ( ivid ) which is associated with

Lk; where p1>p2>. . .>pK and K is the number of priority level in z. Let gk be the

priority row or row (as applicable) corresponding to pk , then the algorithm is as

follows:

Step 1. Initialization:

Set k1

Step 2. Feasibility:

If ib < 0 for i=1,2,…,r then stop. {Solution infeasible}

Step 3. Optimality test:

If k>K, stop.

If all the entries of gk are less than or equal to zero except for the value

corresponding to the column of ivid under consideration and absolute value


of coefficient of deviational variable that are not in the table. go to Step 7.

{All coefficients of priority row gk non positive, so pk is satisfied}.

If the right hand side corresponding to pk is 0, go to step 7

If ib = 0 , for i=1,2,…,r then stop . {Solution optimal}

If all the coefficients of gk row are all less than or equal to zero for all k,

k=1,2,…,K, then stop. {Compromised solution reached}.

Step 4. Entering variable:

(a) Entering variable is the variable with highest positive coefficient (if it is

decision or slack variable) or highest absolute value of coefficient (if

deviational variable) in the row gk. In the case of deviational variable with

highest absolute coefficient, we enter the negative of the deviational

variable (noting that –d- =d+), provided that the deviational variable to enter

is not already in the table.

(b) In case of tie in the coefficients, the entering variable becomes the variable

for which the minimum of the ratio of the right hand side to the positive

coefficients of the columns corresponding to the ties is maximum.

(c) In case of tie in (b), the variable with the least priority will become the

entering variable.

(Priority attached to the entering variable should be placed

alongside with it into the basis).

Step 5. Leaving variable:

i. If 0y is the column corresponding to the entering variable in Step 4, then

the leaving variable is the basic variable with minimum ratio of the right

hand side value to the positive entries in 0y . If 0y is deviational variable

(d+) with highest absolute coefficient, then coefficient of d- will become

coefficient of -d+.

ii. In case of tie in (i), the variable with the smallest right hand side leaves

the basis.

iii. In case of tie in (ii), the variable with the highest priority leaves the


basis.

Step 6. Perform Gauss Jordan row operations to update the table just as in simplex

method of solving LPP. If pk row(s) are still in the basis of the updated

table, go to Step 2.

Step 7. Increment process:

Set k k +1, go to Step 2

REMARKS:

1. The algorithm stops if

i.) The coefficient of the priority rows are all negative or zero for all k

{compromise solution}

ii.) The right hand sides of the priority rows are all zero (optimal solution)

iii.) The priority rule is satisfied and z cannot be achieved further

(compromised solution)

2. The solution is the value of the prioritised deviational variables in the objective

function as appeared in the last iteration tableau.

3. Just as in the method of artificial variables, ensure that a variable of higher or

equal priority that has been previously satisfied does not re-enter the basis,

instead the variable with the next higher coefficient in gk enters.

4. If both d- and d+ (negative and positive deviational variables) from the same goal

constraint are in the achievement function but with different priorities, then the

one with higher priority will be in the basis whereas the lesser one will be placed

alongside with other variables in the non basis.

THE MODIFIED ALGORITHM ON WEIGHTED GOAL PROGRAMMIN

PROBLEMS

In weighted goal programming problem, the algorithm in 4 will be modified to suit

its variant. The variable with the highest coefficient in z is assigned weight w1;

w1> w2 >,…,>wF. and F is the number of weight in z. Relating this to algorithm in 4,

wf ; f=1,2,…,F takes the place of pk and gf takes the place of gk and the algorithm

follows.

The solution becomes the value of the weighted sum of all the deviational variables in


the objective function as appeared in the last iteration tableau; that is the value of the

achievement function becomes a single-valued function.

THE NEW ALGORITHM ON GENERALISED MODEL (GGP)

Recall that in generalised goal programming variant, goals are categorised into

groups, where the goals within each group are of equal importance, but there are

slight differences between the groups in their level of importance such that weight is

used within each group). In this case, the algorithm in section 3.7 will be modified in

order to suit it.

Let pk be the thk priority assigned to deviational variable(s) ( ivid ) which is associated

with group Lk; where p1>p2>. . .>pk and K is the number of priority level in z. Let gk

be the set of rows corresponding to kth priority pk

Also, Let wf be the fth weight assigned to ( ivid ) which is associated with pk in group

Lk; where wf>0 f , then the algorithm is similar to that of 3.7 but with modification

in step 4 as shown below follows;

Step 4. Entering variable:

1. Entering variable is the variable with the highest coefficient in gk.

2. When two or more deviational variables in z have the same priority, then the

entering variable is considered from the deviational variable row(s) with the

highest weight (wf). This will be the variable with highest positive coefficient (if

it is decision or slack variable) or highest absolute value of coefficient (if

deviational variable) in the row(s) corresponding to wf.

3. In case of ties in the coefficient of row(s), then the entering variable is the

variable for which minimum of the ratio of the right hand side to the positive

coefficients of the columns corresponding to the ties is maximum.

4. In case of tie in (3), the variable with the least priority or weight will become the

entering variable (Priority or weight attached to the entering variable should be

placed alongside with it into the basis).

Go through step 5 to 6 and check if w1 is still in the basis. Continue the process

until f=F, then go to step 7.

Then solution becomes a vector of priority level in z.


STEPS BY STEP ILLUSTRATION OF THE PROPOSED ALGORITHM ON

GENERALISED MODEL (GGP)

The problem in (5) below will be solved using the new proposed approach.

Problem 5

minz= p1

2d , p2 (w1 d1- + w2d3

-) where, (w1=3 and w2=1)

s.t

10 x1+12 x2+ d1- -

1d = 2000

6 x1 + 12 x2 +

2d –

2d = 120

8x1 + 4 x2 + d3- _d3

+ = 64

Problem 5 above has two groups. Group one has only one deviational variable, where

as group two has two deviational variable. Within the second group weight “w1 and

w2 ” are used.

Table 5.1: Initial Table of problem 5

x1 x2 d1- d2

- d3- RHS

p2 w1d1

- 10 12 1 0 0 2000

p1d2- 6 12 0 1 0 120

p2 w2d3

- 8 4 0 0 1 64

x2 enters, d2- leaves

Column one represents the variables in the achievement function with priorities and

weights assigned to each of them which forms the bases. Column two and three

represent the coefficients of the decision variables in the goal constraints equation.

Column four to six represent coefficients of deviational variables (-vi) in the goal

constraint equations that appeared in the achievement function. Applying the

algorithm,

Step 1 . Initialization: Here, k=1, r=3, K=2

Set k 1

Step 2. b1, b2, b3>0 {So solution feasible.}


Step 3. gk ={ 6,12,0,1, 0}. Therefore, some entries of gk is non negative, and b2

0 . {So solution not optimal}.

Step 4. max{ 1g }=max{6, 12, 0, 1, 0} =12 which corresponds to x2. So x2 enters

the bases.

Step 5. yo= {gi2}i

min

,0, 2

2

ii

gg

RHS= min {2000/12, 120/12, 64/4}=10

at d2-. Therefore d2

- leaves the bases.

Step 6. Perform Gauss Jordan row operations to update the table and check if

p1 is still in the basis to test for feasibility and optimality. (see tableau

5.2)

Table 5.2: 1st iteration table of problem 5

x1 x2 d1- d2

- d3- RHS

p2 w1d1

- 4 0 1 -1 0 1880

x2 1/2 1 0 1/12 0 10

p2 w2d1

- 6 0 0 -1/3 1 24


Table (5.2) shows that p1 has left the basis, so go to step 7.

Step7. Set k=2, go to Step 2.

Step 2. b1, b3>0 {So solution feasible.}

Step 3. gk ={4, 0, 1, -1, 0,},{6,0, 0, -1/3,1,}. Therefore, some entries of gk is

non negative, and b1 and b2 0 . {So solution not optimal}.

Step 4. Therefore the entering variable will be choosing from the deviational

variable row with the highest weight which is p2

w1d1-. max{{4, 0,

1, -1, 0,}}=4 at g11. So x1 enters the basis .

Step 5. y0= g.1Therefore, min{ 0: 11ig

RHS gi

}= min{1880/4, 20, 24/6} =4. So,

in this case,d3- leaves the bases.


Step 6. Perform the same operation to update the new tableau and check if p2

w1 is still in the basis to test for optimality. (See Table (5.3.

Table 5.3: 2nd iteration table of problem 5

x1 x2

1d d2

- d3- RHS

p2 w1d1 0 0 1 -7/9 -2/3 1864

x2 0 1 0 1/9 -1/12 8

x1 1 0 0 -1/18 1/6 4

d2+ enters

X1 leaves

Table (5.3) shows that p2

w1 is in the basis, so go to step 2.

Step 2. b1 >0 {So solution feasible.}

Step 3. gk = g2 ={0, 0, 1, 9/7 , 3/2 }. Therefore, some entries of gk is

non negative, and b1 0 . {So solution not optimal}.

Step 4. max{0, 0, 1, 9/7 , 3/2 }.=7/9 at –g14 which corresponds to d2+.

Recall that ii dd . Therefore, -(-g14 ) =g14 =7/9. So d2

+ enters the

basis.

Step 5. y0 =-gi4 :{ 0: 44

igRHS g

i} =

4*18,,

7

9*1864=72 which

corresponds to x1. So, x1 leaves and column for d2

+ created.

Step 6. Perform the same operation to update the new tableau and check if p2 is still in the basis to test for feasibility and optimality. (See Table

(5.4).


Table 5.4 : 3rd iteration table of problem 5

x1 x2

1d d2

- d3- d2

+ RHS

p2 w1d1 -14 0 1 0 -3 0 1808

x2 2 1 0 0 1/4 0 16

d2+ 18 0 0 -1 3 1 72

d3+ enters

d1- leaves


w1d1- is still in the basis, so go to step 2.

Step 2 . b1 >0 {So solution feasible.}

Step 3. gk ={-14, 0, 1,0, 3 ,0}. Therefore, some entries of gk is

non negative, and b1 0 . {So solution not optimal}.

Step 4. Max ={-14, 0, 1,0, 3 ,0}=3 which is -d3- =d3

+. So d3+ enters the

basis.

Step 5. y0 = -gi5 : min{ 0: 55

igRHS g

i}=min ,,

31808

=1808/3 which

corresponds to d1-. So, d1

- leaves and a column of is created for d3+

created.

Step 6. Perform the same operation to update the new tableau and check if p

2 w1d1

- is still in the basis to test for feasibility optimality. (See Table

(5.5)).

Table 5.5 : 4th and final iteration table of problem 5

x1 x2

1d

d2- d3

- d2+ d3

+ RHS

d3+ -14/3 0 1/3 0 -1 0 1 1808/3

x2 -3/2 1 1/12 0 0 0 0 500/3

x1 -4 0 1 -1 0 1 0 1880



w1 has left the basis, so go to step 7.

Step7. Set k=3, go to step 2

Step 2. bi >0 {So solution feasible.}

Step 3. k>K=2, so stop.

The current solution is z= p1 (d2-), p2( 3d1

-+ d3-) =0

ILLUSTRATION OF THE ALGORITHM ON GENERALISED MODEL

(WHEN BOTH d- AND d+ (POSITIVE AND NEGATIVE DEVIATIONAL

VARIABLES) FROM THE SAME CONSTRAINT ARE IN THE

ACHIEVEMENT FUNCTION Z)

The problem in (6) below as shown in Journal of operational Res. Soc. Vol 33, No3,

1983, will be solved using the new proposed method.

Problem 6

Min z = p1 d1- , p2d2

-, p3 d3+, p4 d1

+

s.t

2x1 +4x2 +d1- -

1d = 80

8x1 + 10x2 +

2d –

2d = 320

8x1 +6x2 + d3- -d3

+ =240

0,0,,0

iiiii ddddx

Jnl of optl Res. Soc. Vol 33, No3, 1983

Table 6.1: Initial iteration of problem 6

Basis x1 x2 d1- d2

- d3+ d1

+ RHS

p1 d1- 2 4 1 0 0 -1 80

p2 d2- 8 10 0 1 0 0 320

p3d3+ 8 6 0 0 -1 0 240



Column one is the basis which consists of the deviational variables in z with priority

attached to each of them. Here, d1- and d1

+ (positive and negative deviational

variables) from the first constraint are in the achievement function z(i.e p1 d1- and p

4 d1

+). Therefore, the one with highest priority which is

1d is placed on the basis,

whereas the lesser one d1+ is placed alongside with other variables in the non basis.

Column two to six is the matrix of the coefficients of the decision variables from both

goal and rigid constraints and deviational variables that appeared in z.

Row one to three are the coefficients of the variables from the constraint equations

corresponding to p1, p2, p3 ( 321 ,, ggg ) respectively. The variable d1+ which

corresponds to p4 appeared only in the non-basis, since the one with higher priority d1-

has been represented in basis.

Applying the algorithm,

Step 1. Initialization: Here, k=1, r=4, K=4

Set k 1

Step 2 . b1, b2, b3>0 {So solution feasible.}

Step 3. pk =gk ={2,4, 1,1,0,-1}. Some entries of gk is non negative, and b1 0

{So solution not optimal}.

Step 4. Max gk =max{2,4, 1,1,0,-1}=4 which corresponds to x2. So x2 enters the

bases.

Step 5. yo= { 2ig }i

min

,0, 2

2

ii

gg

RHS= min {80/4, 320/10, 240/6}=20.

Therefore d1- leaves the basis.

Step 6. Perform Gauss Jordan row operations to update the table and check if

p1 is still in the basis to test for feasibility and optimality. (see tableau

6.2).

Table 6.2: 1st iteration of problem 6

x1 x2 d1- d2

- d3+ d1

+ RHS

x2 1/2 1 1 0 0 -1/4 20

p2 d2- 3 0 -10/4 1 0 10/4 120

p3d3+ 5 0 -6/4 0 -1 6/4 120

x1 enters, d3+ leaves



Step7. Set k=2, go to Step 2.

Step 2. b2, b3>0 {So solution feasible.}

Step 3. gk ={3,0, -10/4, 1, 0, 10/4}. Some entries of gk is non negative, and b2

0 {So solution not optimal}.

Step 4. Max{3, 0, -10/4, 1, 0, 10/4}= 3. So x1 enters the basis.

Step 5. y0= g.1Therefore, min { 0: 11ig

RHS gi

}= min{40, 40, 24} =24. So, in

this case, d3+ leaves the bases.

Step 6. Perform the same operation to update the new tableau and check if p2

is still in the basis to test for feasibility and optimality. (See Table

(6.3).

Table 6.3: 2nd iteration of problem 6

x1 x2 d1- d2

- d3+ d1

+ RHS

x2 0 1 2/5 0 1/10 -2/5 8

p2 d2- 0 0 -8/5 1 3/5 8/5 48

x1 1 0 -3/10 0 -1/5 3/10 24

d1+ , enters, d2

- leaves

Table (6.3) shows that p2 is still in the basis, so go to step 2.

Step 2. b2>0 {So solution feasible.}

Step 3. pk =gk ={0, 0, -8/5, 1, 3/5,8/5}. Some entries of gk is non negative,

and b2 0 {So solution not optimal}.

Step 4. Max gk = max {0, 0,-8/5, 1, 3/5,8/5}=8/5 which corresponds to d1+.

Therefore, d1+ enters the basis.

Step 5. y0 =gi6 :min{ 0: 66ig

RHS gi

} =

3

10*24,

8

5*48, ={-,30,80}=30

which corresponds to d2-. So, d2

- leaves.

Step 6. Perform the same operation to update the new tableau and check if p2 is

still in the basis to test for feasibility and optimality.


Table 6.4: Last iteration of problem 6

x1 x2 d1- d2

- d3+ d1

+ RHS

x2 0 1 0 ¼ 1/4 0 20

p4 d1+ 0 0 -1 5/8 3/8 1 30

x1 1 0 0 -3/16 -5/16 0 15


Step7. Set k =4.


Step 3. gk ={0,0,-1, 5/8, 3/8, 1}. Some entries of gk is non negative, and b2 0

{So solution not optimal}.

Step 4. Max gk =max {0,0,-1, 5/8, 3/8, 1}=1at which corresponds to d1+. But this

cannot enter for itself to leave. So consider the next higher value which

is 5/8 that corresponds to d2-. But variable with higher priority will not

re-enter for lesser one to leave. So we consider the next higher value

which is 3/8 that corresponds to d3+ with higher priority also. Since there

is no other positive variable in the row, go to 7.

Step7. Set k =5.


Step 3. k>K, so stop.

Therefore, z = 1p

1d , 2p

2d , 3p

3d , 4p

1d = {0,0,0,30}.

SUMMARY AND CONCLUSION

A new approach for solving GLGP is developed, along side with modification of the

existing ones. Difference structure of LGP were enumerated and solved using the new

approach. The new algorithm is efficient.


REFERENCES

[1] Lee, S. M.: Goal Programming for Decision Analysis. Auerbach,

Philadelphia. (1972).

[2] Ignizio J. P. Goal Programming and Extensions. D.C. Heath and company,

Lexington, Massachusetts. (1976).

[3] Olson, D. L.: Comparison of four Goal Programming Algorithms. Journal of

the Operational Research Society Vol. 35, No. 4(1984) pp 347-354.

[4] Schniederjans, M. J. & N. K. Kwak An alternative solution method for goal

programming pro blems: a tutorial. Journal of Operational Research

Society.33 (1982) 247-252.

[5] Orumie, U.C, & D.W.U Ebong .: An alternative method of solving goal pro

gramming problem. Nigerian Journal of operations research. Vol2, no 1

(2011) pp 68-90.

[6] Orumie, U.C, & D.W.U Ebong .: An Efficient Method Of Solving

Lexicographic Linear Goal Programming. International Journal of Scientific and Research Publications, Volume 3. (2013):

[7] Orumie, U.C, & D.W.U Ebong Comparisms of Linear Goal Programming

Algorithm. International Journal of Mathematical methods.2013.

www.iiste.org ISSN 2224-5804 .Vol.3. (2013):

[8] Orumie, U.C, & D.W.U Ebong another Efficient Method of Solving

weighted Linear Goal Programming. Asian Journal of mathematics and Applications Volume , Article ID ama0105, 11 2013): s

[9] Orumie, U.C, & D.W.U Ebong Glorious literature on Linear Goal

programming. American Journal of Operations Research, 2014, 4, 59- 4, 59-71 (2014)

http://www.iiste.org/

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